# how to classify epimorphisms from a subgroup to itself?

Assume $G$,$\hat{G}$ are both free group of rank $n$,and $H$,$\hat{H}$ be their subgroups of index $k$ respectively,$h:H \rightarrow G$, $\hat{h}:\hat{H} \rightarrow\hat{G}$, are two epimorphisms. We call the triple $(H,G,h)$ and $(\hat{H},\hat{G},\hat{h})$ are equivalent if there is a isomorphism $\alpha:G \rightarrow \hat{G}$ satisfies (a)$\alpha(G)=\hat{G}$,$\alpha(H)=\hat{H}$ (b)$\alpha \circ h(x)=\hat{h} \circ \alpha(x)$,$\forall x \in H$ (1)Then I have three questions:(1)when are $(H,G,h)$ and $(\hat{H},\hat{G},\hat{h})$ equivalent?

(2)If we already know $(H,G,h)$ and $(\hat{H},\hat{G},\hat{h})$ are equivalent, how can we find an $\alpha$ ?

(3)For given $n$ and $k$,Can we give a classfy of all those triples in this way?

For all the questions above, I have considered simple the case $n=3$ and founded it's really difficulty for me. I don't know whether this question is trivial or not in Group Theory. If you can give me some useful advices i will be gratiful. Thanks!

• There is no need to call anyone a genius. You should remove that from your question, as it does not add anything of value and, quite the contrary, probably detracts from it. – Mariano Suárez-Álvarez Oct 31 '13 at 3:35
• You've used the notation '$h$' twice in two different ways. Please clarify. – HJRW Oct 31 '13 at 13:53
• I think you need to fix the condition on your isomorphism $\alpha$. First, $h(x) \in \hat{H}$, so you can't apply $\alpha$ to it. Second, $\alpha(x) \in \hat{G}$, so you can't apply $\hat{h}$ to it. – S. Carnahan Oct 31 '13 at 13:54
• To HJRW: thanks, that is my mistake. I have corrected it. – Grub Oct 31 '13 at 14:07
• @HJRW such epimorphisms are called "virtual endomorphism" in the book "self-similar group". It's natural to give a classify of such object. – Grub Nov 1 '13 at 2:04

Here's a naive procedure to answer part (2) of your question. Let $S=(s_1,\ldots,s_n)$ be a minimal generating set for $H$. Using Whitehead's algorithm, say, we can enumerate minimal generating sets $(\hat{s}_i)$ of $\hat{H}$. Now the assignment $s_i\mapsto\hat{s}_i$ gives you a candidate automorphism $\alpha$.
Since $H$ is of finite index, we can check membership of $H$ and $\alpha(H)$, and hence determine whether or not $\alpha(H)=H$. To check that $\alpha\circ h=\hat{h}\circ\alpha$, it suffices to check thison the generators of $H$. This is a finite number of identities, which can be checked using the solution to the word problem in free groups.
• Thanks for your answer! But I think it really not easy to find the solution to the word problem in free groups. In the case $n=3,k=2$ , this need to solve a equation of the following type:$h^{\circ 3} \dots=1$ ,Then I don't know how to deal with it.By question (3) I means that "what is the space of all those triples module the equivalence" – Grub Nov 2 '13 at 10:50